**Space curve:** A space curve is defined as the locus of a point whose cartesian coordinates are the functions of a single variable parameter u, say.

When the curve is not a plane curve, it is said to skew, twisted or tortuous.

**Equation of space curve:** The equation of the space curve at the point u is given by

x = x(u) ————————(i)

where s is any vector of any point on the curve corresponding to the parameter u, say the point P on the curve x = x(u) is generally denoted by P(u) corresponding to the value of parameter u.

Also the equation of the space curve at the point s is given by

x = x(s) ———————–(ii)

where x is any vector of any point on the curve corresponding to the arc length s (say).

**Example:** The following are the equation of space curve

(i) x = x(u)

(ii) x = (cos u, sin u, cu)

**Tangent line to a curve:** The tangent line to a curve at a point P(u) is defined as the limiting position of a straight line L through P(u) and neighbouring point Q(u +du) on the curve as Q approaches P (i.e., Q → P) along the curve.

**Unit tangent vector to a space curve:**

Let P ≡(x, y, z),

Q≡(x +dx, y+dy, z+dz),

= s, = ds, and

= s + ds .

*Let P and Q be two neighbouring points on the curve C whose position vectors are x and x + dx referred to the origin O. Now from ∆OPQ, we have*

= x+dx-x

= dx

## Let A be a fixed point on the space curve and arc AP = s and are AQ = s + ds, so that PQ = ds. Now, unit vector along chord PQ is

Now if the point Q tends to P, then the chord PQ tends to be a tangent at P. Thus we have

Hence unit vector along tangent at P

Hence if the unit tangent vector at P is t, then we have

**Osculating plane (or plane of curvature)**: Let Q, P, R be three consecutive points on a space curve. The plane passing thorough Q, P, R when Q → P and R→P is called the osculating plane at P on the space curve.

**Equation of osculating plane**: Let the equation of the space curve be x = x(u) and Q(u_{2}), P(u_{1}) and R(u_{3}) be three consecutive points on this space curve. Let the equation of the plane passing through Q(u_{2}), P(u_{1}) ,R(u_{3}) be

y(a).a = p,

where a is a constant vector.

So, we have

y(u_{1}).a = p, y(a_{2}).a =p, y(u_{3}).a = p

⟹ y(u_{1}).a –p=0, y(a_{2}).a –p=0, y(u_{3}).a –p=0

Also x= x(u) curve lies on thise plane, so we have

x(u_{1}).a –p=0, x(u_{2}).a –p=0, x(u_{3}).a –p=0

∴ x(u).a = 0 at u_{1}, u_{2}, u_{3}

Suppose that f(u) = x(u).a-p=0

Thus at at u_{1}, u_{2}, u_{3}, we have

f(u_{1})=0, f(u_{2})=0, f(u_{3})=0.

Hence according to Rolle’s theorem, we have

f(v_{1})=0, u_{1}<v_{1}<u_{2}

f(v_{2})=0, u_{1}<v_{2}<u_{2}.

Now f(v) is continuous in [v_{1}, v_{2}] and hence again applying Rolle’s, we get

f(v_{3})=0, v_{1}<v_{3}<v_{2}.

*When the points Q(u _{2}), R(u_{3}) i.e., u_{2}, u_{3} approach u_{1}, then all the points u_{2}, u_{3}, v_{1},v_{2}, v_{3} →u_{1}*

Now the equation of the plane is

y(a). a – p=0 ——————(i)

Also,y(u_{1}) = x(u_{1}).a –p=0 ——————–(ii)

ẏ (u) = ẋ(u_{1}).a =0 —————————(iii)

ӱ (u) = ẍ(u_{1}).a =0 ————————–(iv)

Subtracting (ii) from (i), we get

y(u).a- x(u_{1}).a =0

⟹{ y(u)- x(u_{1})}.a =0 ———————-(v)

Replacing u_{1} by u in (v) , (iii) and (iv), we get

y(u)- x(u)}.a =0, ẋ(u).a =0, ẍ(u).a =0

⟹ (y-x) = 0, ẋ . a =0, ẍ . a =0 where x = x(u)

Since all the vectors (y-x), ẋ, ẍ are perpendicular to the constant vector a, so (y-x), ẋ, ẍ lies on the same plane.

Hence [(y-x) ẋ ẍ] = 0——————(vi)

which is the equation of osculating plane.

## If y = (X, Y, Z) and x = (x, y, z), then substituting these values in (vi), the equation of the osculating plane given by